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Q1. Five years ago the sum of the ages of A & B was 58 years. The difference between B's age 8 years ago and A's age 8 years hence is 16 years. The ratio of the present ages of A & B is:
Q2. What approximate will come in the place of the question mark ‘?’ in the following question?5.78% of 799.94 + ?% of 9.67 = 10.94 × 2.99 + √100 × 2.98
Q3. What approximate value will come in the place of the question mark ‘?’ in the following question?9.99% of 29.906 + 299.84% of 54.908 – 49.86% of 149.59 = ?
Q4. What approximate value should come in place of the question mark (?) in the following question?63.92 ÷ √(255.89) × 24.91% of (2.99)3 = (?)3
Q5. What approximate value should come in the place of the question mark ‘?’ in the following question?728.821/3+√1155.98 + 6.142÷ 2.992+ 1.970=√?
Q6. What approximate value should come in the place of the question mark ‘?’ in the following question?11.11% of 99.17+ 22.22% of 98.87 -9.89% of 100.12 = ?
Q7. In a group of 160 people, 100 people like tennis & 20 people like both cricket & tennis. If each one likes at least one of these two games,then find the ratio between the number of people who like only cricket and the number of people who lượt thích only tennis.
Q8. A certain sum of money is distributed among Ravi, Rahul, và Raj in ratio 8 : 5 : 7 in such a way that giới thiệu of Ravi was Rs. 1000 less than that the sum of mô tả of Rahul và Raj. Find the difference between the shares of Ravi and Raj?
Q9. Radha's expenditure is 20% of her income. If the income for next month is increased by 20%, and the amount of savings remains the same, then find the percentage increase in expenditure of Radha.
Q10. A can vị a piece of work in 22 days while B alone can complete the whole work in 26 days. They work together for 5 days then A left the work và remaining work can done by B alone. In how many days will finish the whole work?
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I want to lớn factorize the polynomial $x^3+y^3+z^3-3xyz$. Using chiaseyhoc.comematica I find that it equals $(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$. But how can I factorize it by hand?


Note that (can be easily seen with rule of Sarrus)$$ eginvmatrix x & y & z \ z và x và y \ y & z và x \ endvmatrix=x^3+y^3+z^3-3xyz$$

On the other hand, it is equal to (if we add to the first row 2 other rows)$$ eginvmatrix x+y+z và x+y+z & x+y+z \ z và x và y \ y & z và x \ endvmatrix=(x+y+z)eginvmatrix 1 và 1 & 1 \ z và x và y \ y & z và x \ endvmatrix=(x+y+z)(x^2+y^2+z^2-xy-xz-yz)$$ just as we wanted. The last equality follows from the expansion of the determinant by first row.

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nói qua
answered Dec 27, 2013 at 15:17

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địa chỉ cửa hàng a bình luận |
eginalignx^3+y^3+z^3-3xyz\&= x^3+y^3+3x^2y+3xy^2+z^3-3xyz-3x^2y-3xy^2\&= (x+y)^3+z^3-3xy(x+y+z)\&= (x+y+z)((x+y)^2+z^2-(x+y)z)-3xy(x+y+z)\&= (x+y+z)(x^2+2xy+y^2+z^2-yz-xz-3xy)\&= (x+y+z)(x^2+y^2+z^2-xy-yz-zx)endalign

chia sẻ
edited Aug 23, 2021 at 21:40

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answered Oct 29, 2013 at 10:23

Yurii Savchuk
Yurii Savchuk
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địa chỉ cửa hàng a comment |
Consider the polynomial $$(lambda - x)(lambda - y)(lambda - z) = lambda^3 - alambda^2+blambda-c ag*1$$We know$$egincasesa = x + y +z\ b = xy + yz + xz \ c = x y zendcases$$ Substitute $x, y, z$ for $lambda$ in $(*1)$ and sum, we get$$x^3 + y^3 + z^3 - a(x^2+y^2+z^2) + b(x+y+z) - 3c = 0$$This is equivalent to$$eginalign x^3+y^3+z^3 - 3xyz= và x^3+y^3+z^3 - 3c\= và a(x^2+y^2+z^2) - b(x+y+z)\= & (x+y+z)(x^2+y^2+z^2 -xy - yz -zx)endalign$$

cốt truyện
answered Oct 29, 2013 at 11:16

achille huiachille hui
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showroom a comment |
Use Newton"s identities:

$p_3=e_1 p_2 - e_2 p_1 + 3e_3$ & so $p_3-3e_3 =e_1 p_2 - e_2 p_1 = p_1(p_2-e_2)$ as required.


$p_1= x+y+z = e_1$

$p_2= x^2+y^2+z^2$

$p_3= x^3+y^3+z^3$

$e_2 = xy + xz + yz$

$e_3 = xyz$

chia sẻ
edited Oct 29, 2013 at 11:31
achille hui
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answered Oct 29, 2013 at 10:29
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add a phản hồi |
A polynomial from $chiaseyhoc.combbQ$ is a polynomial from $chiaseyhoc.combbQ$, so it can be viewed as a polynomial in $z$ with coefficients from the integral domain $chiaseyhoc.combbQ$.$$p(z)=z^3-3xy cdot z +x^3+y^3$$

So we can try our methods to factor a polynomial of degree 3 over an integral domain:If it can be factored then there is a factor of degree $1$, we call it $z-u(x,y)$ and $u(x,y)$ divides the constant term of $p(z)$ which is $x^3+y^3$. The latter is can be factored khổng lồ $(x+y)(x^2-xy+y^2)$ We kiểm tra each of the possible values $(x+y), -(x+y), (x^2-xy+y^2), -(x^2-xy+y^2)$ for $u(x,y)$ & find that only $p(-x-y)=0$. So $z-(-x-y)$ is a factor.


One can use Kronecker"s method

to reduce the factorization of a polynomial of $chiaseyhoc.combbQ$ lớn factoring polynomials in $chiaseyhoc.combbQ$, khổng lồ reduce the factorization of polynomial of $chiaseyhoc.combbQ$ khổng lồ factoring polynomials in $chiaseyhoc.combbQ$to reduce the factorization of polynomial of $chiaseyhoc.combbQ$ to factoring numbers in $chiaseyhoc.combbZ$

This factoring is possible in a finite number of steps but the number of steps may become lớn high for practical purpose.

An integral domain name is a commutative ring with $1$, where the following holds:$$a e 0 land b e 0 implies ab e 0$$For polynomials $f$, $g$, $h$ $in I$ this guarantees:$$f=g cdot h implies extdegree(f)= extdegree(g) + extdegree(h) ag1$$compare this to lớn $chiaseyhoc.combbZ_4$ which is no integral domain and $(2z^2+1)^2 equiv 1$ & so $(2z^n+1) mid 1$. So the polynomial $1$ of degree $0$ has infinitely many divisor.If $I$ is an integral domain $(1)$ guarantees that $z^3+az^2+bz+c in I$ has a linear factor và therefore zero in $I$ if it is not irreduzible.